How do you solve Minimize XOR on LeetCode?

Minimize XOR (LeetCode #2429) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Understanding how XOR works can help in minimizing the result by manipulating bits effectively.

What is the time complexity of Minimize XOR?

The optimal solution for Minimize XOR runs in O(1) time and uses O(1) space. This complexity is constant because we are only iterating through a fixed number of bits (32 bits for integers).

What is the brute force solution for Minimize XOR?

The brute force approach for Minimize XOR is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible integers with the same number of set bits as num2 and calculating their XOR with num1. We then select the integer that yields the smallest XOR value. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Minimize XOR?

Minimize XOR uses the following concepts: Greedy, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Greedy Algorithms.

What are the interview tips for Minimize XOR?

Always explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Be prepared to discuss the properties of XOR and how they can be leveraged in problem-solving. Practice bit manipulation problems, as they often appear in interviews and require a good understanding of binary operations.

What are common mistakes when solving Minimize XOR?

Not considering the bit positions correctly when trying to minimize XOR. Failing to count set bits accurately, leading to incorrect results.

#2429

Minimize XOR

Medium

GreedyBit ManipulationBit ManipulationGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

💡

Intuition

Time O(1)Space O(1)

The optimal approach leverages bit manipulation to construct the integer x directly. By focusing on the bits of num1 and num2, we can efficiently find the integer x that minimizes the XOR without generating all possibilities.

⚙️

Algorithm

6 steps

1Step 1: Count the number of set bits in num2.
2Step 2: Initialize x to 0 and a variable to track remaining set bits.
3Step 3: Iterate through the bits of num1 from the most significant to the least significant bit.
4Step 4: If the current bit in num1 is 0 and we still have set bits left, set that bit in x to 1.
5Step 5: If the current bit in num1 is 1, set that bit in x to 1 if we still have set bits left, otherwise leave it as 0.
6Step 6: If there are still set bits left after processing num1, fill the least significant bits of x with 1s.

solution.py19 lines

1# Full working Python code
2class Solution:
3    def minXor(self, num1: int, num2: int) -> int:
4        set_bits = bin(num2).count('1')
5        x = 0
6        remaining_bits = set_bits
7        for i in range(31, -1, -1):  # Check bits from 31 to 0
8            if (num1 & (1 << i)) == 0 and remaining_bits > 0:
9                x |= (1 << i)  # Set this bit in x
10                remaining_bits -= 1
11            elif (num1 & (1 << i)) != 0 and remaining_bits > 0:
12                x |= (1 << i)  # Set this bit in x
13                remaining_bits -= 1
14        # Fill remaining bits with 1s from the least significant side
15        for i in range(0, 32):
16            if remaining_bits > 0:
17                x |= (1 << i)
18                remaining_bits -= 1
19        return x

ℹ

Complexity note: This complexity is constant because we are only iterating through a fixed number of bits (32 bits for integers).

1Understanding how XOR works can help in minimizing the result by manipulating bits effectively.
2The number of set bits in the target number (num2) is crucial in constructing the optimal solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.